Question

Let S be the set of all sets and let $R=\left\{ \left( A,B \right):A\subset B \right\}$ , i.e. A is a proper subset of B. Show that R is (i) transitive (ii) not reflexive (iii) not symmetric.

Answer 1

Hint:To show that the given relation is transitive find an example such that $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ then $\left( a,c \right)\in R$ for $a,b,c\in S$ . To show that the given relation is not reflexive, find an example such that $\left( a,a \right)\in R$ for all $a\in S$ . To show that the given relation is not symmetric, find an example such that $\left( a,b \right)\in R$ but $\left( b,a \right)\notin R$ for $a,b\in S$ .

Complete step-by-step answer:
Let ‘A’ be a set then Reflexivity, Symmetry and transitivity of a relation on set ‘A’ is defined as follows:
Reflexive relation: - A relation R on a set ‘A’ is said to be reflexive if every element of A is related to itself i.e. for every element say (a) in set A, $\left( a,a \right)\in R$ .
Thus, R on a set ‘A’ is not reflexive if there exists an element $a\in A$ such that $\left( a,a \right)\notin R$.
Symmetric Relation: - A relation R on a set ‘A’ is said to be symmetric relation if $\left( a,b \right)\in R$ then $\left( b,a \right)$must be belong to R. i.e. $\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R$ For all $a,b\in A$.
Transitive relation:- A relation R on ‘A’ is said to be a transitive relation If $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ then $\left( a,c \right)\in R$.
I.e. $\left( a,b \right)\in R$and $\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R$.
Given relation $R=\left\{ \left( A,B \right):A\subset B \right\}$i.e. A is a proper subset of B.
Let A = Set of all natural numbers.
Therefore, $A=\left\{ 1,2,3,4,5,6.............. \right\}$
B= Set of whole numbers.
Therefore, $B=\left\{ 0,1,2,3,4,5,6....... \right\}$
C= Set of all integers.
Therefore, $c=\left\{ -3,-2,-1,0,1,2,3........... \right\}$
Check transitivity: If \[\left( a,b \right)\in R\] and $\left( b,c \right)\in R$ for $a,b,c\in S$ , then $\left( a,c \right)\in R$ , then we can say that R is transitive.
If, \[\left( a,b \right)\in R\] , that means a is a proper subset of b i.e. $a\subset b$ ………(1).
If $\left( b,c \right)\in R$ , that means b is a proper subset of c i.e. $b\subset c$ ………(2).
From (1) and (2) , we can conclude that $a\subset c$ i.e. a is a proper subset of c, Hence $\left( a,c \right)\in R$ .
From the above discussed sets A,B and C, we can see $A\subset B$ and $B\subset C$ because every element of A is present in B and every element of B is there in set C. We can also notice that every element of A is present in set C. Hence $A\subset C$ .
Therefore, R is transitive.
Check Reflexivity: If $\left( a,a \right)\notin R$ , foe $a\in S$ , then R is not reflexive.
Any set can never be the proper subset of itself, because every set is equal to itself i.e. $A\not\subset A$ , because A=A
We can also see in the discussed sets A,B and C. Every set is equal to itself but the given relation requires a set which is a proper subset of another set and proper subsets does not include equal sets.
Therefore, R is not reflexive.
Check symmetricity: If \[\left( a,b \right)\in R\] for $a.b\in S$ and $\left( b,a \right)\notin R$ , Then R is not symmetric.
If \[\left( a,b \right)\in R\] that means a is a proper subset of b i.e. $a\subset b$ .
But, that does not justify that b is also a proper subset of a. That means , b includes a, neither a=b, nor a includes b.
Therefore b is not a subset of a.
$\Rightarrow b$ is not a proper subset of a.
Hence, $\left( b,a \right)\notin R$ .
Therefore, R is not symmetric.

Note: To prove that R is transitive, be careful that if $\left( a,b \right)\in R$and $\left( b,c \right)\in R$ then $\left( a,c \right)\in R$ . If there is no pair such that $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$. Then we don’t need to check, R will always be transitive in this case.